Sect3.2 - ‘ 65:24“): I‘m‘ro +0 DE ' SEC-no.) 3.1...

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Unformatted text preview: ‘ 65:24“): I‘m‘ro +0 DE ' SEC-no.) 3.1 1’” MI?” W .5. (6’3 lélififia} ‘f/LIQ ‘ ‘ V. ‘s (382.01 E‘nd 9h: Wrmska‘ap. _ \ but. K ’v (3) = -21 Hill): 16.31 . .Dokrmme 4M ldrfied. mkrm] Muck W IV? as Wu; 49 W a maul Jute difimaa'hhfle soluhmt ’9) “"Wu' 3“ 1d, * ‘13 = 35” ; yt-2)=2 s‘ g’l~z) =1. Put W8 m3 ‘Hw 3" 413(1):], 4. 3(1):! = 3(2) ’ as 0'» Theorem 32.! 1 n _ 3:25 I' + :1. = .81? y 1"33 (1’33 ((1-0 ' V'W M ’W PC“ . i1“ 30c) 771? \eanc‘HONS pa), 3(1) and a“) W Cow/Mum 14* -ao 4. 1L 4! and [1.x L00. The M40401 Wanna me Delumche 4“ largest ublavaJ ui (duck ~+m Iv? t‘S cerium 4o Lave q web“. hump. ka wed-MA. (ID (1—353" + 13’ +(AI1133 = o 303:0 s‘ 3'11)”. put W; MW 1%,» HM PmU' + $1014 = am! as 0"— ~“woman 3.2.l‘. u _‘_¥-_ l + v . ___.___ = 0. 1d "(1338 + ((1—3 a m0 Pm “37‘” 3‘ The— ?Lmd-tm PHD 4‘8 Wnum: m '00‘1‘3 Md 3‘1. 4 00 . The whrva] watamms m. W‘ POME‘ ‘X..=i,, v: ~ao 4143, The awoken gm ~13 Amhnwus *aOl-‘XLOI 04x43) and 341.4». Th2 miervwl loan-Paws w Mal Pamt,7(,°-i’ ts 0414.3. So we deswzd W41 cs 4% Wd-a‘m a? ~0041¢3 and 0414.3! or - 71m {8 a first o‘rdcr lament ebum‘m: $0le ijra‘hnj j! : ‘ EL X. ( c71- 3), = ctex’ :5 C is: ‘57 3m: m3“ 631, End ‘Hw Wain! set" of Salt:th spcwpved b” MGM 32.5: (2!) 3" +3523 :0 74:0. 753 m We! a? seohm 3.1mm w; 5mm has 'em'shm‘ eoefikuem‘t), let Ida): e“. 7km rte“ + re“ - Zea =0 =7 (rz+rez\=o ‘é r= 1 ar r-=--2_ SD = 1' " V ‘5”) 8 M4 3. m= e, z" som ww. 35. A“ Solo:th econ be wrcflen at W Pam» 30!): (2.8" 4 ctvéz". we wusln 40 W Sold-Ions d6 Was ‘Porm suck 4nd: 3|(10\=l , 7i.(7-.\=O and gang: 0 ,‘ 15:11.3 =1. Tim well be m Mmean Salmons. In W5 Problaq 1.: o. ' goth Bnbfi): an ex +CaC-fcx- and Save, 4" I c\ cu.“ mm = c.+ cz =1, => ¢z=,_c‘ Ufa»: c.f2c,,=o 27 C.'2“‘°I\‘° BJ‘LP “2387‘ + €672” [Conhhudl [Dun gztl3= he‘ 1— the?” and Some 9w- b. ‘3 by: 340) = bfi b7; 0 a, 5;: -b. we»: brawl. 4-» hawks-L =7 55% 3 twin"? 3th.) - 436‘ ~ €672" 5}?) and 34x) Sufisflfl EE. and 44a? «Adda—th and on. Wadi/re. We Mammal soled-1m: 06 Theorem 3.2.5. Section 3.2, Problem 24 Problem. Verify that y1 (t) = e‘t and 3/2 (t) = tet are solutions of the differential equation y” — 2y’ + y = 0. Do they constitute a fundamental set of solutions? Solution. To verify that yl (t) and y2(t) are solutions, one must take the first and second derivatives of both. . y1(t) = 6%, and yflt) = 6“, and y§’(t) = 6‘, so that y§’—2yi+y = et—2et+et = 0. So y1(t) is a solution. y2(t) = tet, y§(t) = e‘ + te‘, and yflt) = 26‘ + tet, so that yg’ -— 23/1 + y = (26‘ + te‘) — 2(6t + tel) + tet = 2et + tet — 26‘ — 2tet + te‘ = 0. So y2(t) is a solution. yl (t) and y2(t) form a fundamental set of solutions if their Wronskian is never zero. The Wronskian is: at tet ‘ et 6‘ + te‘ y1(t) 2/26) yam yam = We +te)_(te)(e) The Wronskian simplifies to (et)(e‘ + tet — tet) == (e?)(et) = 62‘, which is never equal to zero. Since the Wronskian is never equal to zero, y1(t) and y2(t)' form a fundamental set of solutions. _ ...
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This homework help was uploaded on 04/09/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.

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Sect3.2 - ‘ 65:24“): I‘m‘ro +0 DE ' SEC-no.) 3.1...

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